Minimum-value-of-x-2-1-x-2-1-3-is-

Question Number 112624 by Aina Samuel Temidayo last updated on 09/Sep/20

Minimum value of x^{2} + \frac{1}{x^{2} + 1} - 3 is

Answered by MJS_new last updated on 09/Sep/20

x^{2} + 1 + \frac{1}{x^{2} + 1} - 4

x^{2} + 1 = t ⩾ 1

t + \frac{1}{t} - 4

\frac{d}{d t} [t + \frac{1}{t} - 4] = 1 - \frac{1}{t^{2}} = 0 \Rightarrow t = 1 \Rightarrow x^{2} + 1 = 1 \Rightarrow x = 0

\Rightarrow answer is - 2

Answered by ajfour last updated on 09/Sep/20

(x^{2} + 1) + \frac{1}{(x^{2} + 1)} ⩾ 2

\Rightarrow m i n (x^{2} + \frac{1}{x^{2} + 1} - 3) + 4 = 2

\Rightarrow m i n (x^{2} + \frac{1}{x^{2} + 1} - 3) = - 2

Commented by Aina Samuel Temidayo last updated on 09/Sep/20

Thanks .

Answered by 1549442205PVT last updated on 09/Sep/20

P = x^{2} + \frac{1}{x^{2} + 1} - 3 = x^{2} + 1 + \frac{1}{x^{2} + 1} - 4

= {(\sqrt{x^{2} + 1} - \frac{1}{\sqrt{x^{2} + 1}})}^{2} - 2 ⩾ - 2

The equality ocurrs if and only if

\sqrt{x^{2} + 1} = \frac{1}{\sqrt{x^{2} + 1}} \Leftrightarrow x^{2} + 1 = 1 \Leftrightarrow x = 0

Therefore, P_{\min} = - 2 when x = 0